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## Chemical Kinetics Chapter 12

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**Chemical KineticsChapter 12**H2O2 decomposition in an insect H2O2 decomposition catalyzed by MnO2**Objectives**• Understand rates of reaction and the conditions affecting rates. • Derive a rate of reaction, rate constant, and reaction order from experimental data • Use integrated rate laws • Discuss collision theory and the role of activation energy in a reaction • Discuss reaction mechanisms and their effect on rate law**Chemical Kinetics**• We can use thermodynamics to tell if a reaction is product- or reactant-favored. • But this gives us no info on HOW FAST reaction goes from reactants to products. • KINETICS— the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM.**Reaction Mechanisms**• A reaction mechanism is a sequence of events at the molecular level that control the speed and outcome of a reaction. • The reaction mechanism is our goal!**Reaction Rates**Section 12.1 • Reaction rate = change in concentration of a reactant or product with time. • Three “types” of rates • initial rate • average rate • instantaneous rate**Factors Affecting Reaction Rates**• Physical State of the Reactants • Gas, liquid or solid – how molecules are able to interact with each other • Solids react faster when surface area is greater so fine powders react faster than big chunks • Concentration of Reactants • As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. • Temperature • At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. • Catalysts • Speed up reaction by changing mechanism. • Catalysts don’t get used up themselves**Concentrations & Rates**Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) 0.3 M HCl 6 M HCl**Factors Affecting Rates**• Physical state of reactants**Factors Affecting Rates**Catalysts: catalyzed decomp of H2O2 2 H2O2 2 H2O + O2**Factors Affecting Rates**• Temperature Bleach at 54 ˚C Bleach at 22 ˚C**Determining a Reaction Rate**Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye conc with time — can be determined from the plot. Dye Conc Time**Reaction Rates**C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) Average Rate, M/s The average rate of the reaction over each interval is the change in concentration divided by the change in time:**Reaction Rates**C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • Note that the average rate decreases as the reaction proceeds. • This is because as the reaction goes forward, there are fewer collisions between reactant molecules.**Reaction Rates**C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • A plot of concentration vs. time for this reaction yields a curve like this. • The slope of a line tangent to the curve at any point is the instantaneous rate at that time.**Reaction Rates**C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • The reaction slows down with time because the concentration of the reactants decreases.**Rxn Rate =**-[C4H9Cl] t = [C4H9OH] t Reaction Rates and Stoichiometry C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1. • Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.**Reaction Rates**2NO2(g) 2NO(g)+ O2(g) [C4H9Cl] M In this reaction, the concentration of nitrogen dioxide, NO2, was measured at various times, t.**Reaction Rates and Stoichiometry**2NO2(g) 2NO(g) + O2(g) • What if the ratio is not 1:1? • 2NO can be made from 2NO2 consumed, but only 1 O2 is produced. • Read as: the rate of consumption of NO2 is the same as the rate of production of NO. This is because their coefficients are the same.**Reaction Rates**But since the coefficient for oxygen is ½ of the other two, it’s rate is of production rate is half as fast. Or 2 x rate O2 = rate of NO**Reaction Rates**• Reaction Rate and Stoichiometry summary: • For the reaction • C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • we know • In general for • aA + bBcC + dD • Rxn**Reaction Rates Practice**• What is the relative rate of disappearance of the reactants and relative rate of appearance of the products for each reaction? • 2O3(g) 3O2(g) • 2HOF(g) 2HF(g) + O2(g) • In the synthesis of ammonia, if Δ[H2]/Δt =-4.5 x 10-4mol/L•min, what is rate with respect to NH3? wrt N2? N2(g) + 3H2(g) 2NH3(g)**Concentrations & Rates**2NO2(g) 2NO(g) + O2(g) Rate of reaction is proportional to [NO2] We express this as a RATE LAW Rate of reaction = k [NO2]n where k = rate constant k is independent of conc. but increases with T n is the order of the reactant**Concentrations, Rates, & Rate Laws**In general, for a A + b B x X with a catalyst C Rate = k [A]m[B]n[C]p The exponents m, n, and p: • sum of m, n and p are the reaction order • can be 0, 1, 2 or fractions like 3/2 • must be determined by experiment! They are not simply related to stoichiometry!**Ch. 12.3 - Interpreting Rate Laws**Determining Rate by inspection (quick method): Rate = k [A]m[B]n[C]p • If m = 1, rxn. is 1st order in A Rate = k [A]1 If [A] doubles, then rate goes up by factor of _2_ • If m = 2, rxn. is 2nd order in A. Rate = k [A]2 Doubling [A] increases rate by ___4_____ • If m = 0, rxn. is zero order. Rate = k [A]0 If [A] doubles, rate _Stays the same_**The method of Initial Rates**• This method requires that a reaction be run several times. • The initial concentrations of the reactants are varied. • The reaction rate is measured just after the reactants are mixed. • Eliminates the effect of the reverse reaction.**Deriving Rate Laws**Derive rate law and k for CH3CHO(g) CH4(g) + CO(g) from experimental data for rate of disappearance of CH3CHO Expt. [CH3CHO] Disappear of CH3CHO (mol/L) (mol/L•sec) 1 0.10 0.020 2 0.20 0.081 3 0.30 0.182 4 0.40 0.318**Deriving Rate Laws**Determination of Rate by inspection: Here the rate goes up by __4___ when initial conc. doubles. Therefore, we say this reaction is ________2nd_________ order. So, the Rate of rxn = k [CH3CHO]2 Now determine the value of k. Use expt. #3 data— 0.182 mol/L•s = k (0.30 mol/L)2 k = 2.0 (L / mol•s) Using k you can calc. rate at other values of [CH3CHO] at same T.**Deriving Rate Laws**Determination of Rate by rigorous math: Rate = k[A]n Note: For easier math, put the faster rate in the numerator. 4=2n n=2**Determination of Rate Law Practice**For the reaction BrO3- + 5Br- + 6H+ 3Br2 + 3H2O • The general form of the Rate Law is Rate = k[BrO3-]n[Br-]m[H+]p • We use experimental data to determine the values of n, m, and p • We will choose trials that vary only one of the concentrations to determine the order based upon that reactant.**Determination of Rate Law Practice**Initial concentrations (M) BrO3- • Determine the rate law wrt each reactant • Determine the rate constant, k (with units) • Determine the overall order of reaction Br- H+ Rate (M/s) 0.10 0.10 0.10 0.8 x 10-3 0.20 0.10 0.10 1.6 x 10-3 0.20 0.20 0.10 3.2 x 10-3 0.10 0.10 0.20 3.2 x 10-3**Ch. 12.4 – Integrated Rate Law (Concvs time)**What is concentration of reactant as function of time? Consider FIRST ORDER REACTIONS The rate law is And the units of k are sec-1.**Integrated Rate Law**• Expresses the reaction concentration as a function of time. • Form of the equation depends on the order of the rate law (differential). • Changes Rate = D[A]nDt • We will only work with n=0, 1, and 2**Concentration/Time Relations**Integrating - (∆ [A] / ∆ time) = k [A], we get ln is natural logarithm [A]0at time = 0 [A] / [A]0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law.**Concentration/Time Relations**Sucrose decomposes to simpler sugars Rate of disappearance of sucrose = k [sucrose] If k = 0.21 hr-1 and [sucrose] = 0.010 M How long to drop 90% (to 0.0010 M)? Glucose**Concentration/Time RelationsRate of disappear of sucrose = k**[sucrose], k = 0.21 hr-1.If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M? Use the first order integrated rate law ln (0.100) = - 2.3 = - (0.21 hr-1)(time) time = 11 hours**Using the Integrated Rate Law**The integrated rate law suggests a way to tell the order based on experiment. 2 N2O5(g) 4 NO2(g) + O2(g) Time (min) [N2O5] (M) ln [N2O5] 0 1.00 0 1.0 0.705 -0.35 2.0 0.497 -0.70 5.0 0.173 -1.75 Rate = k [N2O5]**Using the Integrated Rate Law**2 N2O5(g) 4 NO2(g) + O2(g) Rate = k [N2O5] Plot of ln [N2O5] vs. time is a straight line! Data of conc. vs. time plot do not fit straight line.**Using the Integrated Rate Law**Plot of ln [N2O5] vs. time is a straight line! Eqn. for straight line: y = mx + b All 1st order reactions have straight line plot for ln [A] vs. time. ln[N2O5 ] = -kt + ln[N2O5]0 conc at rate constconc at time t =-slope time=0**Half-Life**HALF-LIFE is the time it takes for 1/2 a sample is disappear. For 1st order reactions, the concept of HALF-LIFE is especially useful. See Active Figure 15.9**Half-Life**• Reaction is 1st order decomposition of H2O2.**Half-Life**• Reaction after 1 half-life. • 1/2 of the reactant has been consumed and 1/2 remains.**Half-Life**• After 2 half-lives 1/4 of the reactant remains.**Half-Life**• A 3 half-lives 1/8 of the reactant remains.**Half-Life**• After 4 half-lives 1/16 of the reactant remains.**Half-Life**Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme products Rate of disappear of sugar = k[sugar] k = 3.3 x 10-4 sec-1 What is the half-life of this reaction?**Half-Life**Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the half-life of this reaction? Solution [A] / [A]0 = fraction remaining when t = t1/2 then fraction remaining = _______ Therefore, ln (1/2) = - k · t1/2 - 0.693 = - k · t1/2 t1/2 = 0.693 / k So, for sugar, t1/2 = 0.693 / k = 2100 sec = 35 min NOTE: For a first-order process, the half-life does not depend on [A]0.**Half-Life**Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)? Solution 2 hr and 20 min = 4 half-lives Half-life Time Elapsed Mass Left 1st 35 min 2.50 g 2nd 70 1.25 g 3rd 105 0.625 g 4th 140 0.313 g**Second Order**• Rate = -Δ[A]/Δt = k[A]2 • integrated rate law • 1/[A] = kt + 1/[A]0 • y= 1/[A] m = k • x= t b = 1/[A]0 • A straight line if 1/[A] vs t is plotted • Knowing k and [A]0 you can calculate [A] at any time t**NO2(g)**NO (g) + 1/2 O2(g) Determining rxn order The decomposition of NO2 at 300°C is described by the equation and yields these data: